Graduate Does QFT specify particle propagation?

[friend]

It does not. You cannot count vacuum fluctuations.

Just a moment. Isn't that what the cosmological constant is all about - the amount of quantum fluctuations in a certain volume of space?

If so, then consider that an empty box in a boosted frame is shrunk in the direction of motion. So what are we to think of the space inside the box. Does the box shrink but the space inside it does not? Then if the space inside the boosted box feels the same cosmological constant in a smaller space, then what can that mean except that a still observer would see the same amount of vacuum fluctuations inside a smaller volume?

 

[mfb]

Isn't that what the cosmological constant is all about - the amount of quantum fluctuations in a certain volume of space?

No.

If so, then consider that an empty box in a boosted frame is shrunk in the direction of motion. So what are we to think of the space inside the box. Does the box shrink but the space inside it does not? Then if the space inside the boosted box feels the same cosmological constant in a smaller space, then what can that mean except that a still observer would see the same amount of vacuum fluctuations inside a smaller volume?

Those questions do not make sense.

 

[friend]

No. Those questions do not make sense.

Anything and everything inside the empty boosted box would appear squeezed in the direction of motion, including the subatomic particles, and I assume any vacuum fluctuations that affect those squeezed subatomic particles. I recall an animation of particles colliding in the LHC. It showed the (what, point particles?) flattened out like two pancakes as they approached each other. If quantum fluctuations had any effect at all on these moving particles, then shouldn't we assume they'd be flattened out as well?

 

[mfb]

A vacuum in a box does not have a velocity. It is a vacuum.

If quantum fluctuations had any effect at all on these moving particles, then shouldn't we assume they'd be flattened out as well?

No, that does not make sense. How often do I have to repeat that?

 

[FieldTheorist]

Perhaps vacuum fluctuations look the same at any constant speed. I know acceleration does funny things to the vacuum fluctuations. But if every inertial observer sees every other inertial observer's vacuum fluctuations to be the same as his, then I have my answer. Just as a fast moving yard stick appears shorter to those not moving, must the vacuum fluctuations of a fast mover be different than for a still mover? If not, then no, QFT cannot explain kinetic energy; that's just put in by hand as the energy of the input states.

I would honestly suggest that you stick to the conceptual objects of "quantum particle" and "quantum field," before you traipse off into discussions using (very antiquated) terminology that you almost certainly don't understand. Vacuum fluctuations (like "holes" in the "Dirac sea") are a way of giving a picture to certain subcalculations of an observable, but the picture is pretty meaningless and just words to dress around a calculation.

If you want to take QFT seriously, you need to start by treating the quantum field as fundamental, learn the language you need to describe it, see how the Fock space structure of free QFTs leads to an infinite tower of states of N identical particles, and discover what the fundamental, physical observables of the theory are.

A fast moving yard stick appears shorter to those not moving. So must the vacuum fluctuations seen by a fast moving observe be different than the vacuum fluctuations seen by a motionless observer? Would we say that the vacuum fluctuations seen by a fast moving observer are squeezed WRT us? I take it that there is so many quantum fluctuation per meter for any observer, and the fast moving meter is squeezed. So wouldn't we say that the fast moving observer must be experiencing a squeezed version of the vacuum fluctuations. Does this make any sense?

Let's take a step back. The vacuum is the state of zero particles. (If you accelerate, you'll see a bath of particles, but it's a coordinate artifact. Let's assume we're in a Lorentz frame.) When I say "fluctuations in the quantum field" what I mean, to be clear, is exciting a particle state. If I turn on a source, that will excite the vacuum, and generate particles. Meaning you will no longer be in the vacuum state.

I think I'm getting closer to the right language. A photon on average will encounter so many vacuum fluctuations per meter. The meter is defined by the propagation of light in a vacuum. A meter stick traveling close to the speed of light will appear shorter because ALL processes are squeezed in the direction of motion. Does this mean we still observers see their vacuum fluctuations squeezed as well in the direction of their motion? Since all processes are defined by propagation through the vacuum, And their meter sticks appear shorter, does that mean that their vacuum appears shorter as well?

No, I don't see the statement here. You seem like you need to spend more time reading about QFT before you can ask full, coherent questions on QFT. I recommend reading Tong's intro to QFT, it's free and you can find it with a google search. If you make it through the first two chapters, I think you'll be able to answer many of your own conceptual confusions. Zee's QFT in a Nut Shell might also be a useful read.

 

[friend]

Thank you for your advice. But what I'm hearing from this is basically, "Shut up and calculate". Although, you are being nice about it. I have no commitment at this time to a particular interpretation. I'm just here to gain understanding. I think that requires some visualization of what's going on in the math. And the only visualization technique for quantum theory seems to be that of virtual particles, unless, of course, you know of a better way to visualize what's going on.

 

[Vanadium 50]

Let me be more direct - even if that means being less nice. You are dismissing a correct view in favor of an incorrect view because you like it better. This not science.

 

[friend]

Let me be more direct - even if that means being less nice. You are dismissing a correct view in favor of an incorrect view because you like it better. This not science.

You are attributing a commitment on my part based only on my asking questions. I think this thread is done.

 

[mfb]

I think this thread is done.

Me too.